<span>A rocket in its simplest form is a chamber enclosing a gas under pressure. A small opening at one end of the chamber allows the gas to escape, and in doing so provides a thrust that propels the rocket in the opposite direction. A good example of this is a balloon. Air inside a balloon is compressed by the balloon's rubber walls. The air pushes back so that the inward and outward pressing forces are balanced. When the nozzle is released, air escapes through it and the balloon is propelled in the opposite direction.</span>
What it looks to be that you found in A was the "initial"...b/c the question asks:
<span>"how much energy does the electron have 'initially' in the n=4 excited state?" </span>
<span>"final" would be where it 'finally' ends up at, ie. its last stop...as for this question...the 'ground state' as in its lowest energy level. </span>
The answer comes to: <span>−1.36×10^−19 J</span>
You use the same equation for the second part as for part a.
<span>just have to subract the 2 as in the only diff for part 2 is that you use 1squared rather than 4squared & subract "final -initial" & you should get -2.05*10^-18 as your answer. </span>
In both scenarios, the position - time graph will be a linear graph, since the speed is constant, so your position is moving at a consistent pace.
Answer:
(C) 2P
Explanation:
Ideal gas law states:
PV = nRT
n (the number of moles) and R (ideal gas constant) are constant, so we can say:
(PV / T) before = (PV / T) after
Chamber X starts at pressure P, volume V, and temperature T. At equilibrium, the pressure is Px, the volume is Vx, and temperature 3T.
PV / T = Px Vx / 3T
Chamber Y starts at pressure P, volume V, and temperature T. At equilibrium, the pressure is Py, the volume is Vy, and temperature T.
PV / T = Py Vy / T
Substituting and simplifying:
Px Vx / 3T = Py Vy / T
Px Vx / 3 = Py Vy
Since the chambers are at equilibrium, Px = Py:
Vx / 3 = Vy
Vx = 3 Vy
The total volume is the same as before, so:
Vx + Vy = 2V
Substituting:
(3 Vy) + Vy = 2V
4 Vy = 2V
Vy = V / 2
Now if we substitute into our equation for chamber Y:
PV / T = Py (V/2) / T
PV = Py (V/2)
Py = 2P
The pressure in chamber Y (and chamber X) doubles at equilibrium.
To solve this problem we will apply the concepts related to the calculation of the surface, volume and error through the differentiation of the formulas given for the calculation of these values in a circle. Our values given at the beginning are


The radius then would be

And

PART A ) For the Surface Area we have that,

Deriving we have that the change in the Area is equivalent to the maximum error, therefore

Maximum error:


The relative error is that between the value of the Area and the maximum error, therefore:


PART B) For the volume we repeat the same process but now with the formula for the calculation of the volume in a sphere, so


Therefore the Maximum Error would be,



Replacing the value for the radius


And the relative Error


